Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There is a "folk theorem" (alternatively, a fun and easy exercise) which asserts that a 2D TQFT is the same as a commutative Frobenius algebra. Now, to every compact oriented manifold $X$ we can associate a natural Frobenius algebra, namely the cohomology ring $H^\ast(X)$ with the Poincare duality pairing. Thus to every compact oriented manifold $X$ we can associate a 2D TQFT.

Is this a coincidence? Is there any reason we might have expected this TQFT to pop up?

When $X$ is a compact symplectic manifold, perhaps the appearance of the Frobenius algebra can be explained by the fact that the quantum cohomology of $X$, which comes from the A-twisted sigma-model with target $X$, becomes the ordinary cohomology of $X$ upon passing to the "large volume limit".

But for a general compact oriented $X$? I don't see how we might interpret the appearance of the Frobenius algebra in some quantum-field-theoretic way. Maybe there is an explanation via Morse homology?

share|improve this question
add comment

2 Answers 2

up vote 7 down vote accepted

These 2D TQFTs do not come from extended theories (unless X is discrete). I interpret this as saying that these theories are non-local (in the 2D bordism) and so you will have trouble interpreting them in a traditional QFT framework. You will have to do something funny and non-local, like squashing your circles to points and surfaces to graphs, as in the Cohen work mentioned by Tim.

share|improve this answer
1  
Hi Chris: That is very interesting. To be precise, are you saying that the 2-1 TQFT coming from any non-discrete manifold cannot come from a 2-1-0 TQFT? Is this perhaps somewhere in your thesis? –  Kevin H. Lin Dec 30 '09 at 15:01
4  
Hi Kevin. Yes, that's right. This is Corollary 4.6.15 in my thesis (on page 211). For any 0-1-2 TQFT over a ring the commutative algebra associated to the circle must be separable and projective. Over a field this implies it must be semi-simple, and over Z it is equivalent to being a finite direct sum of copies of Z (as algebras!). In particular, this can only happen if the cohomology is concentrated in degree zero. This statement holds in the framed TQFT setting, and so is true for all variations of 2D TQFTs: oriented, unoriented, spin, with G-bunlde, etc. –  Chris Schommer-Pries Dec 30 '09 at 15:24
1  
That's super awesome. –  Kevin H. Lin Dec 30 '09 at 16:10
add comment

There is indeed a Morse homology explanation; and, in the symplectic case, it's a degeneration of the Hamiltonian Floer cohomology picture. In a nutshell, you degenerate the surfaces to graphs, and then use a different Morse function for each edge. This has been explored (e.g.) by Ralph Cohen, initially in a paper with Betz and more recently with Norbury:

http://arxiv.org/pdf/math/0509681v1.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.